Finding a basis of an infinite dimensional vector space with a given vector












2












$begingroup$


If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






linear-algebra set-theory axiom-of-choice







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



    First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



    My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



    P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






    linear-algebra set-theory axiom-of-choice







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



      First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



      My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



      P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






      linear-algebra set-theory axiom-of-choice







      share|cite|improve this question











      $endgroup$




      If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



      First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



      My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



      P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






      linear-algebra set-theory axiom-of-choice




      linear-algebra set-theory axiom-of-choice






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Robert Shore

      1,32815




      1,32815










      asked 2 hours ago









      JustDroppedInJustDroppedIn

      2,054420




      2,054420






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
            $$
            v = a + k_2a_2 + cdots + k_na_n .
            $$



            Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3119673%2ffinding-a-basis-of-an-infinite-dimensional-vector-space-with-a-given-vector%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






                  share|cite|improve this answer











                  $endgroup$



                  Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  Robert ShoreRobert Shore

                  1,32815




                  1,32815























                      1












                      $begingroup$

                      Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                      $$
                      v = a + k_2a_2 + cdots + k_na_n .
                      $$



                      Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                        $$
                        v = a + k_2a_2 + cdots + k_na_n .
                        $$



                        Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                          $$
                          v = a + k_2a_2 + cdots + k_na_n .
                          $$



                          Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                          share|cite|improve this answer









                          $endgroup$



                          Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                          $$
                          v = a + k_2a_2 + cdots + k_na_n .
                          $$



                          Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Ethan BolkerEthan Bolker

                          43.5k551116




                          43.5k551116






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3119673%2ffinding-a-basis-of-an-infinite-dimensional-vector-space-with-a-given-vector%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Knooppunt Holsloot

                              Image
                              Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                              Read more

                              Altaar (religie)

                              Image
                              Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                              Read more

                              Gregoriusmis

                              Image
                              Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                              Read more